Let $p_n$ be the first $n$ primes in some order. For which $n, n \ge 2$, does there exist a $k$ such that $$\prod_{i=1}^{k} p_i=\left(\prod_{j=k+1}^{n} p_j\right )+1$$ has a solution?
I’ve determined that there exists a solution for $n\in \{2,3,4\}$, but I haven’t found any solutions after that (searching manually). I came across this organically after noticing that $14,15$, which are $1$ apart, are the product of all the primes up to $7$. I noticed that this concept is similar to the idea of Ruth-Aaron numbers (https://youtu.be/aCq04N9it8U), wherein the sums of the prime factors are equal, but didn’t see a specific way to apply that concept.
If $P_k$ is the product of the first $k$ primes, you're asking whether $P_k$ has a divisor $d$ (which of course will be the product of some subset of these primes) such that $P_k = d (d+1)$. Now the equation $x(x+1)=P_k$ has solutions $$ x = \frac{-1 \pm \sqrt{1+4 P_k}}{2}$$ We get a positive integer solution (with $+$) if and only if $1 + 4 P_k$ is a square. So the question is, when is $1 + 4 P_k$ a square? It is true for $k = 1$ (if you allow the empty product being $1$), $ 2, 3, 4, 7$ and no others up to $10000$, the corresponding $d$ being $1, 2, 5, 14$ and $714$ respectively. That is, $$ \eqalign{2 &= 1 \cdot 2 \cr 6 &= 2 \cdot 3 \cr 30 &= 5 \cdot 6 = 5 \cdot (2 \cdot 3) \cr 210 &= 14 \cdot 15 = (2 \cdot 7) \cdot (3 \cdot 5)\cr 510510 &= 714 \cdot 715 = (2 \cdot 3 \cdot 7 \cdot 17) \cdot (5 \cdot 11 \cdot 13)\cr} $$
EDIT: See OEIS sequence A161620, according to which there are no other solutions with $k < \pi(10^{11})$.